LOGARITHMS
Definitions
If ax is equal to N, then x is called the logarithm of N to be base a.
In this case, we write loga N = x.
Thus the logarithm of a number to a given base is the index to which the base should be raised to get the
number.
Example 1:
As 16 = 24, log2 16 = 4
As 16 = 42, log4 16 = 2.
Example 2:
As
Example 3:
We find the value of log2 8. Now, if log2 8 = x, then x = 3.
Example 4:
What is log1/2 4?
Exercise 1:
Express the following in the logarithm form:
1
1
= 2–2, log2
= –2.
4
4
(a) 0.001 = 10-3
(d) 5 = 5
1 2
 4  answer is –2.
2
(b) 64 = 82
1
(e) 30 = 1
c
-4
(g) a = b
(j) 7 = 343
0
(h) 3 =
1
81
(c) 64 = 26
(f) 81 = 274/3
4
1
(i)   = 0.0625
2 
1
3
Laws of Logarithms
Example 5:
23 + 2 = 2 3 22
i.e. 25 = 8  4
Thus, log2 (8  4) = 5
=3+2
= log2 8 + log2 4.
Similarly log5 (125  125) = log5 125 + log5 125
This gives:
Law 1:
Loga (xy) = loga x + loga y
Example 6:
23  22 = 2 3 – 2
 log2 (8  4) = 1
=3–2
= log2 8 – log2 4
Similarly log3 (81  9) = log3 81 – log3 9
=4–2=2
This gives:
=3+3=6
x
 = loga x – loga y
y
Law 2:
loga 
Example 7:
43 = 64
Now, 4 = 22
 log2 4 = 2
Now, 64 = 43

64 = (2  2)3 = 23  23
= 23 + 3 = 26
 log2 64 = 6
=32
= 3 log2 4
3
i.e. log2 4 = 3 log2 4.
Exercise 2:
Expand loga xn = loga (x  x ….  x) … (n times)
Law 3:
loga xr = r loga x
Cor. log
an
x = 1/n loga x
Change of Base
Loga n  logb n . loga b
Logb n =
loga n
loga b
Exercise 3:
Expand loga [ x6 (x  2)4]
Exercise 4:
Expand log4
Exercise 5:
Show that loga 7 + 2 loga 5 + 5 loga 10 – loga 28 – loga 10 – 6 loga 5 – 2 loga 2 = 0.
Exercise 6:
If x2 + y2 = 11xy show that loga
Exercise 7:
Solve
Exercise 8:
Solve log(3x + 2) – log(x – 1) = log4
Exercise 9:
If log V – log4 = log  + 3 log r – log 3, find the formula for V in terms of  and r.
Exercise 10:
Find the formula for V if
Exercise 11:
Show that p(q – 1) = q, if log p + log q = log (p + q)
5
log a 100
loga 10
x7
x-y
3
=
loga x + loga y
2
=x
logV - logh
2
= log r +
1
2
log 
a+b
Exercise 12:
Show that a = b if 2 log
Exercise 13:
Given 2 log x + 1 = log 250, find (a) x (b) log 2x.
Exercise 14:
Given that log 2 = 0.301 and log 3 = 0.4771, evaluate (a) log 1.2 (b) log 12.5
2
= log a + log b.
Logarithms to the base ‘10’ are called Common logarithms.
Characteristic and Mantissa
We know that
100 = 1  log10 1 = 0
101 = 10  log10 10 = 1
102 = 100  log10 100 = 2
103 = 1000  log10 1000 = 3
104 = 10,000  log10 10,000 = 4
and so on.
similarly,
10-1 = 1/10 = .1  log10 0.1 = -1
10-2 = 1/10 = .01  log10.01 = -2
10-3 = 1/1000 = .001  log10.001 = -3
10-4 = 1/10000-.0001  log10.0001 = -4
You see that log 1 = 0, log 10 = 1 and log 100 = 2
But what about the logarithms of numbers lying between 1 and 10 or 10 and 100? From the pattern it is clear
that the logarithms of numbers lying between 1 and 10 i.e. having one digit whole numbers will be between
‘0’ and ‘1’. They will be greater than ‘0’ and less than i.e. they will be 0. (……) some decimal fraction.
When using common logarithms base 10 is not written. Thus log10 10 is simply written as log 10 and so on.
Similarly logarithm of numbers lying between 10 and 100 i.e. numbers having 2-digit whole numbers will be
1. (…..) (some decimal fraction).
And logarithm of numbers having 3 digit whole numbers, i.e. which lie between 100 and 1000 will be 2. (…..)
(some decimal fraction).
Now, coming to numbers which are less than 1, i.e. 0.1, .01, .001 etc. their logarithms will be between -1 and
-2, -2 and -3, -3and -4 respectively and so on. Thus they shall also have a number and decimal part as in
case of logarithm of a numbers greater than 1.
Thus, the logarithm of a number has a whole number part as well as a decimal part.
The whole number part of the number is called Characteristics.
As you have seen that the characteristic of the logarithm of a number is positive if the number is greater than
1 and negative if the number is less than 1 (i.e. a fraction).
So, the characteristic can be positive as well as negative.
The fractional or decimal part of the logarithm of a number is called Mantissa.
The mantissa of the logarithm of a number is always positive and is read from Logarithm tables.
Illustration 1:
Solution:
Find the antilogarithm of –3.8134.
As – 3.8134 = -3-0.8134,
we have -3.8134 = - 3 - (1 – 0.1866)
= 4.1866.
 antilog (-3.8134) = antilog 4 .1866 = 1.537  10-4.
The tables can be used to simply products.
 2.812 
Illustration 2:
Find x where x =
Solution:
log x = 3 log 2.812 +
3
× 12.912
6.134 ×  4.913 
4
1
1
log 12.912  log 6.134 – 4 log 4.913
2
2
1
1
= 3(0.4490) +
(1.1109) - log 6.134 – 4log 4.913
2
2
= - 1.25705 = 1.2571
= 2 .7429
Thus x = antilog 2 .7429 = 5.533  10–2
= 005533.
Exercise 15:
Use the logarithm tables to find the value of
Exercise 16:
Given that log x = 2 log 5.87 -
Exercise 17:
Find x if
1
2
(a) (log x)2 – 3 log x + 2 = 0
0.816
61.82
to three significant digits.
log 0.839, find x correct to 3 significant figures.
(b) (log x)2 + log x = 0
ANSWERS TO EXERCISES
Exercise 2:
n loga x
Exercise 3:
10 loga x – 4 loga 2
Exercise 4:
7/5 log4 x
Exercise 7:
2
Exercise 8:
7
Exercise 9:
V=
Exercise 10:
(r2h)
Exercise 13:
a) 5 b) 1
Exercise 14:
a) 0.0791 b) 1.097
Exercise 15:
0.8049
Exercise 16:
37.6
Exercise 17:
(a) 10,100
4
3
r3
(b) 0.1, 1
ASSIGNMENTS
SUBJECTIVE
LEVEL – I
1.
Given that 2 logx + 1/2 log y = 1, express y in terms of x.
2.
Show that log
3.
If a =
ab 1
(log a + log b) implies that a2 + b2 = 6ab
2 2
4.
b
then show that log (a + b) = log a + log b
b 1
Solve:
(a) log 72 – log20 = 2log 3 + x + 3 log2 + log5
(b) log (3x + 12) – log (2x + 1) = log5
5.
Given that log 3 = 0.48 and log 7 = 0.84, find log
6.
Given that log 3 = 0.4771 and log 7 = 0.8457, find the value of log 2⅓
7.
Given that log m = x + y and log n = x – y, express the value of log m2 n in terms of x and y.
8.
9.
10.
0.03
0.7
 10x 
Given that log x = m + n and log y = m – n, express the value of log  2  in terms of m and n.
 y 
32
Given that log 2 = 0.301 and log 3 = 0.477, find the value of log
30
Given that log x = a, log y = b,
(a) Evaluate 10a – 1in terms of x
(b) Evaluate 102b in terms of y
(c) If log P = 2a – b, express P in terms of x and y.
LEVEL – II
1.
Show that n log m + m log n – 1 = 0, implies that nm .mn = 10
2.
Write the logarithmic equation for
mm
1
(a) F = G 12 2
(b) E = mc2
2
r
(c) T = 2
3.
Show that : log 125 = 3(1 – log 2).
4.
If T = 2
5.
Use common logarithms to find x if x = log2 log2 32.2
1
, find T when  = 3.1416, l = 1.32 and g = 9.809.
g
l
g
T
g
ANSWERS
SUBJECTIVE
LEVEL – I
1.
Y = 100/x4
4.
(a) 2 (b) 1
5.
2.64
7.
3x + y
8.
1 + 3nm
9.
0.1028
10.
(a)
x
10
(b) y2
(c)
x2
y
LEVEL – II
4.
2.305
5.
2.321
6.
0.3686